The simulation of a train moving loads on a continuous beam with seven elastic supports located at same level (II)
THE SIMULATION OF A TRAIN MOVING LOADS ON A CONTINUOUS BEAM WITH SEVEN ELASTIC SUPPORTS LOCATED AT THE SAME LEVEL (II) Cornel MARIN Affiliation: Valahia University of Targoviste Post Adress: Bulevard Carol I, Nr. 2, Targoviste, Jud. Dambovita, 130024, ROMANIA Email:
[email protected] Abstract. This paper continue with the simulation (I) of shear efforts T(x), bending efforts M(x), and displacements w(x) depending on the train moving loads (known) for a continuous beam with seven elastic supports (proportionally with reaction forces) and allows numerical simulation by changing the value of the variable distance from the beginning of the beam. Key words: continuous beams, elastic supports, moving simulation
1.
Introduction
One considers a train moving load (y=variable) on a continuous beam supported by seven elastic supports at the same level located at distances r 1 , r 2 , … r 7 from the beginning of the beam (figure 1). The bending stiffness of continuous beam EI is constant. The continuous beam subjected to bending with train moving load, forces P 1 , P 2 , P 3 placed at varying distance y from the beginning of the beam .
The displacements w1, w2, … w7 are proportionally with reaction forces V1, V2, … V7: V V V (1) w1 = 1 ; w2 = 2 ; .... w7 = 7 k k k where k is the stiffness of elastic bearings (same for all elastic bearings).
d3 d2 y=d1
P1
O
1 r1
V1 r2
P2
P3
2
3
4
5
6
7
V2
V3
V4
V5
V6
V7
x
r3 r4 r5 r6 z
r7 Figure 1. Continuous beam model supported by seven elastic supports at the same level
2.
Calculus of reaction forces
One will determine the bearing reactions forces V 1 , V 2 , ... V 7 and plot using MathCAD program, the diagrams of the shear effort T z (x), bending effort M iy (x),
slope ϕ y (x) and displacement w(x) depending on the variable distance y. In order to determine the reactions V 1 , V 2 , ... V 7 on write two equilibrium conditions between the exterior loads P 1 , P 2 , P 3 and the reactions:
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The simulation of a train moving loads on a continuous beam with seven elastic supports located at same level (II)
∑ Fzs = V1 + V2 + V3 + V4 + V5 + V6 + V7
(2)
∑ M 7 s = V1 ⋅ ( r7 − r1 ) + V2 ⋅ ( r7 − r2 ) + V3 ⋅ ( r7 − r3 ) + (3) + V4 ⋅ ( r7 − r4 ) + V5 ⋅ ( r7 − r5 ) + V6 ⋅ ( r7 − r6 )
where:
∑ Fzs = P1 + P2 + P3
The other five relations will result by replacing the origin parameters (w 0 and ϕ 0 ) in the displacement equations from the supports: 1 W ( r1 ) = w1 EI 1 W ( r2 ) = w2 + EI 1 W ( r3 ) = w3 + EI 1 W ( r4 ) = w4 + EI 1 W ( r5 ) = w5 + EI 1 W ( r6 ) = w6 + EI 1 + W ( r7 ) = w7 EI
+ ϕ 0 ⋅ r3 + ϕ 0 ⋅ r4 + ϕ 0 ⋅ r5 + ϕ 0 ⋅ r6 + ϕ 0 ⋅ r7
k3 ( r1 ) = k31 , k3 ( r2 ) = k32 ... k4 ( r1 ) = k41 , k4 ( r2 ) = k42 ...
(9b)
k5 ( r1 ) = k51 ; k5 ( r2 ) = k52 ... . k6 ( r1 ) = k61 , k6 ( r2 ) = k62 ... k7 ( r1 ) = k71 , k7 ( r2 ) = k72 ...
... k p1 ( r1 ) = k p11 , k p1 ( r2 ) = k p12 ...and so on k p 2 ( r1 ) = k p 21 , k p 2 ( r2 ) = k p 22 ...
(9c)
k p 3 ( r1 ) = k p 31 , k p 3 ( r2 ) = k p 32 ...
+ ϕ 0 ⋅ r1 + + ϕ 0 ⋅ r2
k1( r2 ) = k12 ... and so on
k 2 ( r1 ) = k 21 , k 2 ( r2 ) = k 22 ...
(4)
is the exterior loading train on Oz axis; ∑ M 7 s = P1 ⋅ ( r7 − d1 ) + P2 ⋅ ( r7 − d 2 ) + P3 ⋅ ( r7 − d 3 ) (5) the moment of exterior forces on the Oy axis considered as passing through support no. 7.
w0 w 0 w 0 w0 w0 w0 w0
k1( r1 ) = k11 ;
Important Note: kij = 0 , if i ≥ j
(6)
(10)
The expression (8) becomes to simply: W ( x ) = P1 ⋅ k p1 ( x ) + P2 ⋅ k p 2 ( x ) + P3 ⋅ k p 3 ( x ) − V1 ⋅ k1 ( x ) − V2 ⋅ k 2 ( x ) − − V3 ⋅ k 3 ( x ) − V4 ⋅ k 4 ( x ) − V5 ⋅ k 5 ( x ) − V6 ⋅ k6 ( x ) − V7 ⋅ k7 ( x )
(11)
If we eliminate w0 and ϕ0 in the first-second and second-third equations (6) on obtain: w − w2 W ( r2 ) − W ( r1 ) w − w1 W ( r3 ) − W ( r2 ) − EI 2 = − EI 3 r2 − r1 r2 − r1 r3 − r2 r3 − r2
(12)
where: - the displacements corresponding to rigid supports are proportionally with corresponding reaction forces:
w1 =
V V1 V ; w2 = 2 ; .... w7 = 7 k k k
(7)
- the second integral of bending moments according to relation (8) for the particular case of figure 3 will be: ( x − d 3 )3 ( x − d1 )3 ( x − d 2 )3 + P2 ⋅ Φ(x − d 2 ) ⋅ + P3 ⋅ Φ(x − d 3 ) ⋅ − 6 6 6 ( x − r3 )3 ( x − r4 )3 ( x − r1 )3 ( x − r2 )3 − V1 ⋅ Φ(x − r1 ) ⋅ − V2 ⋅ Φ(x − r2 ) ⋅ − V3 ⋅ Φ(x − r3 ) ⋅ − V4 ⋅ Φ(x − r4 ) ⋅ − 6 6 6 6 ( x − r5 )3 ( x − r6 )3 ( x − r7 )3 − V5 ⋅ Φ(x − r5 ) ⋅ − V6 ⋅ Φ(x − r6 ) ⋅ − V7 ⋅ Φ(x − r7 ) ⋅ 6 6 6
W ( x ) = P1 ⋅ Φ(x − d1 ) ⋅
Similarly are obtained the other four equations: W ( r3 ) − W ( r2 ) − EI r3 − r2 W ( r4 ) − W ( r3 ) − EI r4 − r3 W ( r5 ) − W ( r4 ) − EI r5 − r4 W ( r6 ) − W ( r5 ) − EI r6 − r5
( x − r1 )3 ( x − r2 )3 ; k 2 ( x ) = Φ( x − r2 ) ⋅ 6 6 ( x − r3 )3 ( x − r4 )3 k3 ( x ) = Φ(x − r3 ) ⋅ ; k4 ( x ) = Φ(x − r4 ) ⋅ 6 6 k1 ( x ) = Φ( x − r1 ) ⋅
( x − r5 )3 ( x − r6 )3 ; k6 ( x ) = Φ(x − r6 ) ⋅ ; 6 6 3 ( x − r7 ) k7 ( x ) = Φ(x − r7 ) ⋅ 6 ( x − d 1 )3 k p1 ( x ) = Φ (x − d 1 ) ⋅ ; 6 ( x − d 2 )3 k p 2 ( x ) = Φ(x − d 2 ) ⋅ 6 ( x − d 3 )3 k p 3 ( x ) = Φ(x − d 3 ) ⋅ 6 k5 ( x ) = Φ(x − r5 ) ⋅
62
w4 − w3 W ( r5 ) − W ( r4 ) w − w4 = − EI 5 r4 − r3 r5 − r4 r5 − r4
(13)
w5 − w4 W ( r6 ) − W ( r5 ) w − w5 = − EI 6 r5 − r4 r6 − r5 r6 − r5 w6 − w5 W ( r7 ) − W ( r6 ) w − w6 = − EI 7 r6 − r5 r7 − r6 r7 − r6
On denote:
(8) On denote:
w3 − w2 W ( r4 ) − W ( r3 ) w − w3 = − EI 4 r3 − r2 r4 − r3 r4 − r3
α 121 =
k − k12 k12 − k11 , α 132 = 13 , ... an so on r2 − r1 r3 − r2
α 221 =
k − k 22 k 22 − k 21 ,α 232 = 23 , ... r2 − r1 r3 − r2
(14)
...
(9a)
α 6 21 =
k62 − k61 k − k62 ,α 632 = 63 , ... r2 − r1 r3 − r2
d 12 = r2 − r1 ; d 23 = r3 − r2 ; d 34 = r4 − r3 ; d 45 = r5 − r4 ; d 56 = r6 − r5 ; d 67 = r7 − r6 .
(15)
Separating the terms containing the unknowns reaction forces V1, V2 , … V7 from equations (18) and (19), and taking account at equations (9), (10) we obtain the matricial form as it follows: The Romanian Review Precision Mechanics, Optics & Mechatronics, 2014, No. 46
The simulation of a train moving loads on a continuous beam with seven elastic supports located at same level (II)
A⋅ X = B 1 1 1 1 1 − − − − − r r r r r r r r r r5 7 1 7 2 7 3 7 4 7 EI 1 1 EI 1 EI 1 − − − + + − α α α α 0 0 2 2 1 1 21 32 21 32 k d12 d 23 k d 23 k d12 EI 1 1 EI 1 EI 1 + − α 132 − α 143 α 232 − α 243 − α 332 − α 343 + 0 k d 23 d 34 k d 34 k d 23 EI 1 1 EI 1 EI 1 + − α 143 − α 154 α 243 − α 254 α 343 − α 354 − α 443 − α 454 + k d 34 d 45 k d 45 k d 43 EI 1 1 EI 1 + α 154 − α 165 α 254 − α 265 α 354 − α 365 α 443 − α 454 − α 543 − α 554 + k d 45 d 56 k d 54 EI 1 α 165 − α 176 α 265 − α 276 α 365 − α 376 α 465 − α 476 α 565 − α 576 − k d65
1 r7 − r6 0 0 0
EI 1 − k d 56 EI 1 1 + α 665 − α 676 + k d 56 d67
∑ Fzs M 7 s ∑ k p 22 − k p 21 k p 23 − k p 22 k p 32 − k p 31 k p 33 − k p 32 k p12 − k p11 k p13 − k p12 − ⋅ + − ⋅ + − ⋅ P P P 3 1 r −r 2 r −r − − − − r r r r r r r r 2 1 3 2 2 1 3 2 2 1 3 2 k p13 − k p12 k p14 − k p13 − k p 22 k p 24 − k p 23 k k − k p 32 k p 34 − k p 33 ⋅ P2 + p 33 ⋅ P3 ⋅ P1 + p 23 − − − r4 − r3 r4 − r3 r4 − r3 r3 − r2 r3 − r2 r3 − r2 = k − k p13 k p15 − k p14 k p 24 − k p 23 k p 25 − k p 24 k p 34 − k p 33 k p 35 − k p 34 p14 ⋅ P1 + ⋅ P2 + ⋅ P3 − − − r −r r −r r4 − r3 − − r5 − r4 r r r r 4 3 5 4 4 3 5 4 − − k k k k k k k k k k k k − − − − p 14 p 16 p 15 p 24 p 26 p 25 p 34 p 36 p 35 p15 ⋅ P1 + p 25 ⋅ P2 + p 35 ⋅ P3 − − − r −r r −r r5 − r4 r6 − r5 r6 − r5 r6 − r5 5 4 5 4 k k k k k k k k k k k k − − − − − − p 15 p 17 p 16 p 25 p 27 p 26 p 35 p 37 p 36 p16 ⋅ P1 + p 26 ⋅ P2 + p 36 ⋅ P3 − − − r − r r7 − r6 r7 − r6 r7 − r6 6 5 r6 − r5 r6 − r5
The matricial relation (15) will be used in MathCAD replacing the following numerical values of the parameters:
r 1 =10 m; r 2 =20 m; r 3 =30 m; r 4 =40 m; r 5 =50m; r 6 =60 m; r 7 =70 m;
y=d 1 - variable parameter;
P 1 =P 2 =P 3 =25 kN/m;
d 2 -d 1 =4 m ; d 3 -d 2 =4 m
EI=1000 kNm2
k=100 kN/m
0 V1 V 2 0 V3 ⋅ V = 4 0 V5 V6 V7 0 EI 1 − k d67 1 0
(15)
The simulation of function: shear effort T z (x) , bending moment M iy (x) and displacement w(x) for seven numerical values of the parameter y, are presented in Figure 2 … Figure 10.
100
80
60
40 T ( x)
20
− M ( x) Axa ( x)
0
− EIw( x)
100
− 20
− 40
− 60
− 80
− 100
0
10
20
30
40
50
60
70
80
x
Figure 2. The diagrams for particular value y=10m (Mmax=66,39 kNm)
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The simulation of a train moving loads on a continuous beam with seven elastic supports located at same level (II)
100
80
60
40 T ( x)
20
− M ( x) Axa ( x)
0
− EIw( x)
100
− 20
− 40
− 60
− 80
− 100
0
10
20
30
40
50
60
70
80
x
Figure 3. The diagrams for particular value y=12m (Mmax=65,347 kNm)
100
80
60
40 T ( x)
20
− M ( x) Axa ( x)
0
− EIw( x)
100
− 20
− 40
− 60
− 80
− 100
0
10
20
30
40
50
60
70
80
x
Figure 4. The diagrams for particular value y=14m (Mmax=63,883 kNm) 100
80
60
40 T ( x)
20
− M ( x) Axa ( x)
0
− EIw( x)
100
− 20
− 40
− 60
− 80
− 100
0
10
20
30
40
50
60
70
x
Figure 5. The diagrams for particular value y=16m (Mmax=43,037kNm)
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The simulation of a train moving loads on a continuous beam with seven elastic supports located at same level (II) 100
80
60
40 T ( x)
20
− M ( x) Axa ( x)
0
− EIw( x)
100
− 20
− 40
− 60
− 80
− 100
0
10
20
30
40
50
60
70
80
x
Figure 6. The diagrams for particular value y=18m (Mmax=48,126 kNm)
100
80
60
40 T ( x)
20
− M ( x) Axa ( x)
0
− EIw( x)
100
− 20
− 40
− 60
− 80
− 100
0
10
20
30
40
50
60
70
80
x
Figure 7. The diagrams for particular value y=20m (Mmax=56,093 kNm)
100
80
60
40 T ( x)
20
− M ( x) Axa ( x)
0
− EIw( x)
100
− 20
− 40
− 60
− 80
− 100
0
10
20
30
40
50
60
70
80
x
Figure 8. The diagrams for particular value y=22m (Mmax=56,432 kNm) The Romanian Review Precision Mechanics, Optics & Mechatronics, 2014, No. 46
65
The simulation of a train moving loads on a continuous beam with seven elastic supports located at same level (II) 100
80
60
40 T ( x)
20
− M ( x) Axa ( x)
0
− EIw( x)
100
− 20
− 40
− 60
− 80
− 100
10
0
20
30
40
50
60
80
70
x
Figure 9. The diagrams for particular value y=24m (Mmax=49,012 kNm) 100
80
60
40 T ( x)
20
− M ( x) Axa ( x)
0
− EIw( x)
100
− 20
− 40
− 60
− 80
− 100
0
10
20
30
40
50
60
70
80
x
Figure 10. The diagrams for particular value y=25m (Mmax=56,432 kNm) 3. Conclusions
4. References
•
1.
•
• • •
•
66
The diagrams express of functions of shear force Tz(x), bending moment Miy(x), transversal crosssection rotations ϕy(x) and displacements w(x), is very easy to write and allows numerical simulation by changing the value of the variable y Using the MathCAD step function this method can be easily generalized for a larger number of supports due to the symmetry of the matrices A and B presented above; This method can be generalized for the case of continuous beams by n≥7 supports at same level by simply imposing the conditions: k→∞ on obtain the case of beam with seven rigid supports located at same level (I); by comparing similarly cases (I) and (II) can be observed that the maximum bending moment Mmax in this case is only positive (corresponding to forces P1, P2 or P3) ; this case a corresponding behavior of a continuous beam located on elastic discrete soil.
2. 3.
4.
5.
6.
C. Marin, Rezistenţa materialelor şi elemente de teoria elasticităţii, Targoviste Editura Biblioteca, 2006 C. Marin, Aplicaţii ale teoriei elasticităţii în inginerie, Targoviste Editura Biblioteca, 2007 C. Marin, A. Marin, Metoda analitica pentru trasarea diagramelor de eforturi în barele drepte, Bucuresti, SIMEC 2006, C. Marin, A. Marin, Metoda analitica pentru calculul deplasărilor şi rotirilor barelor drepte supuse la încovoiere, Bucuresti, SIMEC 2006, pp.87-92 C. Marin, Matricial method used for reaction forces computation in the case of continous beams supported by stiff bearings (I) MECAHITECH’14 Bucharest, 2014. C. Marin, Matricial method used for reaction forces computation in the case of continous beams supported by elastic bearings (II) MECAHITECH’14 Bucharest, 2014.
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